We are given a set \( A = \{1, 2, 3, \ldots, n\} \) and asked to find the value of \( n \) for which the probability that a randomly selected mapping \( f: A \to A \) is injective is \( \frac{3}{32} \).
An injective function (also known as one-to-one) means that different elements in set \( A \) must map to different elements in set \( A \). Therefore, for the function to be injective, all elements must be mapped uniquely.
The total number of functions from \( A \to A \) is \( n^n \), because each of the \( n \) elements in the domain \( A \) has \( n \) choices in the codomain \( A \).
The number of injective mappings is the number of ways to arrange \( n \) distinct elements, which is \( n! \) (factorial of \( n \)).
The probability that a randomly chosen mapping is injective is given by:
\(\frac{n!}{n^n}\)
We are given that this probability is \( \frac{3}{32} \). Therefore, we have the equation:
\(\frac{n!}{n^n} = \frac{3}{32}\)
To solve for \( n \), we will compare the values of \(\frac{n!}{n^n}\) by testing successive integers.
- For \( n = 3 \):
- We have: \( n! = 3! = 6 \)
- And: \( n^n = 3^3 = 27 \)
- So, \(\frac{3!}{3^3} = \frac{6}{27} = \frac{2}{9}\) (not equal to \(\frac{3}{32}\))
- For \( n = 4 \):
- We have: \( n! = 4! = 24 \)
- And: \( n^n = 4^4 = 256 \)
- So, \(\frac{4!}{4^4} = \frac{24}{256} = \frac{3}{32}\)
Thus, the value of \( n = 4 \) satisfies the condition that the probability is \(\frac{3}{32}\).
Therefore, the correct answer is \( n = 4 \) and the correct option is \( \boxed{4} \).
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